Available at: http://www.pmf.ni.ac.rs/filomat
New Weighted Inequalities for Higher Order
Derivatives and Applications
Samet Erdena, Mehmet Zeki Sarikayab, Huseyin Budakb
aDepartment of Mathematics, Faculty of Science, Bartın University, Bartın-Turkey
bDepartment of Mathematics, Faculty of Science and Arts, D ¨uzce University, Konuralp Campus, D ¨uzce-TURKEY
Abstract. We establish a new Ostrowski type inequality for (n+ 1)-times differentiable mappings which are bounded. Then, some new inequalities of Hermite-Hadamard type are obtained for functions whose (n+ 1) th derivatives in absolute value are convex. Spacial cases of these inequalities reduce some well known inequalities. With the help of obtained inequalities, we give applications for the kth−moment of random variables.
1. Introduction
In 1938, Ostrowski established the integral inequality which is one of the fundemental inequalitıes of mathematics as follows (see, [20]):
Let f : [a, b]→ R be a differentiable mapping on (a, b) whose derivative f0
: (a, b)→ R is bounded on (a, b), i.e., f 0 ∞= sup t∈(a,b) f 0 (t)
< ∞. Then, the inequality holds: f (x) − 1 b − a b Z a f (t)dt ≤ 1 4 + x −a+b2 2 (b − a)2 (b − a) f 0 ∞ (1)
for all x ∈ [a, b]. The constant 1
4 is the best possible.
The following inequality is well known in the literature as the Hermite-Hadamard integral inequality (see, [7]): f a+ b 2 ! ≤ 1 b − a Z b a f (x)dx ≤ f (a)+ f (b) 2 (2)
where f : I ⊂ R → R is a convex function on the interval I of real numbers and a, b ∈ I with a < b.
2010 Mathematics Subject Classification. Primary 26D07; Secondary 26D15, 60E15.
Keywords. Hermite-Hadamard inequality; Ostrowski inequality; Midpoint inequality; Convex function; Random variables. Received: 21 November 2016; Revised: 21 October 2018; Accepted: 15 November 2018
Communicated by Dragan S. Djordjevi´c
Email addresses: [email protected] (Samet Erden), [email protected] (Mehmet Zeki Sarikaya), [email protected] (Huseyin Budak)
Inequalities (1) and (2) have wide applications in numerical analysis and in the theory of some special means; estimating error bounds for some special means, some mid-point, trapezoidal and Simpson rules and other quadrature rules, etc. Hence, inequality (1) and (2) have attracted considerable attention and interest from mathematicans and researchers. Now, we give some inequalities related to (1) and (2) which were proved in recent years (see, [7], [8], [11], [21], [24], [26]).
In [8], Cerone et.al. proved the following inequalities of Ostrowski type and Hadamard type respectively.
Theorem 1.1. Let f : [a, b] → R be a twice differentiable mapping on (a, b) and f00: (a, b) → R is bounded, i.e.,
f 00 ∞ = sup t∈(a,b) f 00 (t)
< ∞. Then we have the inequality: f (x) − 1 b − a Z b a f (t)dt − x −a+ b 2 ! f0 (x) ≤ 1 24(b − a) 2+1 2 x − a+ b 2 !2 f 00 ∞ ≤ (b − a) 2 6 f 00 ∞
for all x ∈ [a, b].
Corollary 1.2. Under the assumptions of Theorem 1.1, we have the mid-point inequality:
f a+ b 2 ! − 1 b − a Z b a f (x)dx ≤ (b − a) 2 24 f 00 ∞ (3)
In [11], Kırmacı proved the following results connected with the left part of (2).
Theorem 1.3. Let f : I◦ ⊆ R → R be a differentiable mapping on I◦, a, b ∈ I◦ with a< b. If f 0 is convex on [a, b], then we have 1 b − a b Z a f(t) dt − f a+ b 2 ! ≤b − a 8 f 0 (a) + f 0 (b) . (4)
Theorem 1.4. Let f : I◦ ⊆ R → R be a differentiable mapping on I◦, a, b ∈ I◦
with a< b, and let p > 1. If the mapping f 0 p
p−1 is convex on [a, b], then we have
1 b − a b Z a f(t) dt − f a+ b 2 ! (5) ≤ b − a 16 4 p+ 1 !1p 3 f 0 (a) p p−1 + f 0 (b) p p−1 p−1 p + f 0 (a) p p−1 + 3 f 0 (b) p p−1 p−1 p . Sarikaya et. al. pointed out some inequalities in [24], as follows:
Theorem 1.5. Let f : I ⊂ R → R be a differentiable function on I◦
, with f00∈ L1[a, b] . If f 00 is convex on [a, b], then f a+ b 2 ! − 1 b − a Z b a f (x)dx ≤ (b − a) 2 24 f 00 (a) + f 00 (b) 2 . (6)
Theorem 1.6. Let f : I ⊂ R → R be twice differentiable function on I◦
such that f00∈ L1[a, b] where a, b ∈ I, a < b, If f 00 q
is convex on [a, b], q > 1, then f a+ b 2 ! − 1 b − a Z b a f (x)dx ≤ (b − a) 2 8 2p+ 11p f 00 (a) q + f 00 (b) q 2 1 q . (7)
In a recent paper, in [4], Barnett and Dragomir obtained a variety of bounds for the variance and expected value of a continuous random variable whose p.d.f. is defined over a finite interval base on the identity:
b Z a (t − m1)2 f(t) dt+ (m1−b) (m1−a)= b Z a (t − a) (t − b) f (t) dt where m1= b R a u f(u) du.
In recent years, researchers have studied some integral inequalities by using n-times differentiable functions. For example, Authors gave some Ostrowski type inequalities for mappings whose nth derivatives are bounded in [6] and [29]. Sofo established integral inequalities on Lpnorm in [27]. In [22] and [23], the
authors deduced midpoint and trapezoidal formulas for n-times differentiable mappings, respectively. In [1], [2], [9], [18] and [28], researchers obtained some integral inequalities for functions whose obsolute value of nth derivatives are convex, s−convex, m−convex and (α, m) −convex. Kechriniotis and Theodorou proved some integral inequalities via n-times differentiable functions and gave some applications for probability density function in [10]. In [16], [17] and [19], Latif and Dragomir established Hermite-Hadamard type inequalities for n-times differentiable.
In this study, first of all, we derive an identity for (n+ 1)-times differentiable functions. Then, some weighted integral inequalities are obtained by using this identity. Some results presented in earlier works related to these inequalities are given. Finally, some applications for random variable whose probability density functions are bounded and their derivatives in absolute are convex on the interval of real numbers.
2. Some inequalities for the moments
In order to prove weighted integral inequalities, we need the following lemma:
Lemma 2.1. Let f : I ⊂ R → R be (n + 1) −times differentiable function on I◦, a, b ∈ I◦ with a < b, f(n+1) is absolutely continuous on [a, b] and let w : [a, b] → R be nonnegative and continuous on [a, b]. Then the following equality holds: n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt= b Z a Pw(x, t) f(n+1)(t) dt (8) where n ∈ N, Mk(x) is defined by Mk(x)= b R a (u − x)kw(u) du, k = 0, 1, 2, ... and Pw(x, t) := 1 n! t R a (u − t)nw(u) du, a ≤ t < x 1 n! t R b (u − t)nw(u) du, x ≤ t ≤ b. (9)
Proof. By integration by parts, we have b Z a Pw(x, t) f(n+1)(t) dt = 1 n! x Z a t Z a (u − t)nw(u) du f(n+1)(t) dt+ 1 n! b Z x t Z b (u − t)nw(u) du f(n+1)(t) dt = 1 n! b Z a (u − x)nw(u) du f(n)(x)+ 1 (n − 1)! x Z a t Z a (u − t)n−1w(u) du f(n)(t) dt + 1 (n − 1)! b Z x t Z b (u − t)n−1w(u) du f(n)(t) dt.
By integration by parts n−times, we get
b Z a Pw(x, t) f(n+1)(t) dt = Mn(x) n! f (n)(x)+Mn−1(x) (n − 1)! f (n−1)(x)+ ... +M2(x) 2! f 00 (x) +M1(x) f0(x)+ M0(x) f (x) − b Z a w(t) f (t) dt
which is the required identity in (8). Hence, the proof is completed.
We establish a new inequality for functions whose (n+ 1) −th derivatives are bounded
Theorem 2.2. Suppose that all the assumptions of Lemma 2.1 hold. Additionally, we assume that f(n+1): (a, b) → R
is bounded, i.e., f(n+1) ∞= sup t∈(a,b) f(n+1)(t)
< ∞, then we have the inequality n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt (10) ≤ f (n+1) [a,b],∞ (n+ 1)! ×
Mn+1(x) ,if n is an odd number
b R x (u − x)n+1w(u)du − x R a (u − x)n+1w(u)du , if n is an even number for all x ∈ [a, b] .
can write n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt ≤ 1 n! x Z a t Z a (t − u)nw(u)du f(n+1)(t) dt+ 1 n! b Z x b Z t (u − t)nw(u)du f(n+1)(t) dt ≤ f (n+1) [a,x],∞ n! x Z a t Z a (t − u)nw(u)du dt+ f (n+1) [x,b],∞ n! b Z x b Z t (u − t)nw(u)du dt.
By using the change of order of integration and the fact that n is an odd number, we get n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt ≤ f (n+1) [a,x],∞ (n+ 1)! x Z a (x − u)n+1w(u)du+ f (n+1) [x,b],∞ (n+ 1)! b Z x (u − x)n+1w(u)du ≤ f(n+1) [a,b],∞ (n+ 1)! Mn+1(x). Hence, the proof is completed.
Remark 2.3. If we choose n= 1 in Theorem 2.2, then we obtain
M1(x) f 0 (x)+ M0(x) f (x) − b Z a w(t) f (t) dt ≤ f 00 [a,b],∞ 2 M2(x). which was given by Sarikaya and Yaldiz in [25].
Remark 2.4. If we choose w(u)= 1 in Theorem 2.2, then we have the inequality n X k=0 (b − x)k+1+ (−1)k(x − a)k+1 (k+ 1)! f (k)(x) − b Z a f(t) dt ≤ f(n+1) [a,b],∞ (n+ 2)! h (b − x)n+2+ (x − a)n+2i .
for all n ≥ 0. This inequality was proved by Cerone et al. in [6].
Remark 2.5. If we take w(u)= 1 and n = 0 in Theorem 2.2, then we get the clasical Ostrowski inequality.
Remark 2.6. If we take w(u)= 1 and n = 1 in Theorem 2.2, then the Theorem 2.2 reduces to the Theorem 1.1 which is proved by Cerone et.al. in [8].
Remark 2.7. If we choose w(u)= 1 and x = a+b2 in Theorem 2.2, then we have the inequality n X k=0 (b − a)k+1h1+ (−1)ki 2k+1(k+ 1)! f(k) a+ b 2 ! − b Z a f(t) dt ≤ f (n+1) ∞(b − a) n+2 2n+1(n+ 2)! .
for all n ≥ 0. This inequality was proved by Cerone et al. in [6].
Remark 2.8. If we take w(u)= 1, x = a+b2 and n= 1 in Theorem 2.2, then the inequality (10) becomes the inequality (3) which was given by Cerone et.al. in [8].
Now, we give an inequality for mappings whose absolute value of (n+ 1) −th derivatives are convex.
Theorem 2.9. Suppose that all the assumptions of Lemma 2.1 hold. If f
(n+1)
is convex on [a, b], then, for all x ∈ [a, b] , the following inequality holds:
n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt (11) ≤ kwk[a,b],∞ (n+ 1)! (b − a) " (b − a)(x − a) n+2 n+ 2 + (b − x)n+3−(x − a)n+3 n+ 3 ! f (n+1) (a) + (x − a)n+3−(b − x)n+3 n+ 3 + (b − a) (b − x)n+2 n+ 2 ! f (n+1) (b) # where kwk∞ = sup t∈[a,b] |w(t)|.
Proof. By taking absolute value of (8) and using the boundedness of mapping w, we find that n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt ≤ kwk[a,x],∞ (n+ 1)! x Z a (t − a)n+1 f (n+1) (t) dt+ kwk[x,b],∞ (n+ 1)! b Z x (b − t)n+1 f (n+1) (t) dt. Since f (n+1)
(t) is convex on [a, b] = [a, x] ∪ [x, b] f(n+1) b − t b − aa+ t − a b − ab ! ≤ b − t b − a f (n+1) (a) + t − a b − a f (n+1) (b) . (12)
Utilizing the inequality (12), we write n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt (13) ≤ kwk[a,x],∞ (n+ 1)! (b − a) f (n+1) (a) x Z a (t − a)n+1(b − t) dt+ f (n+1) (b) x Z a (t − a)n+2dt + kwk[x,b],∞ (n+ 1)! (b − a) f (n+1) (a) b Z x (b − t)n+2dt+ f (n+1) (b) b Z x (b − t)n+1(t − a) dt .
If we calculate the above four inetgrals and also substitute the results in (13), because of kwk[a,x],∞, kwk[x,b],∞≤
Remark 2.10. Under the same assumptions of Theorem 2.9 with n= 0, then the following inequality holds: f(x) b Z a w(t) dt − b Z a w(t) f (t) dt ≤ kwk[a,b],∞ (b − a) " (b − a)(x − a) 2 2 + (b − x)3−(x − a)3 3 ! f 0 (a) + (x − a) 3− (b − x)3 3 + (b − a) (b − x)2 2 ! f 0 (b) #
which is ”weighted Ostrowski” inequality provided that f
0 is convex on [a, b] . This inequality was given by
Sarikaya and Erden in [26].
Remark 2.11. Under the same assumptions of Theorem 2.9 with n= 0 and x = a+b2 , then the following inequality hols: f a+ b 2 !Zb a w(t) dt − b Z a w(t) f (t) dt ≤ kwk[a,b],∞(b − a)2 4 f 0 (a) + f 0 (b) 2 which is ”weighted mid-point” inequality provided that
f
0
is convex on [a, b] . This inequality was given by Sarikaya and Erden in [26].
Remark 2.12. If we choose n= 1 in Theorem 2.9, then we obtain
M1(x) f 0 (x)+ M0(x) f (x) − b Z a w(t) f (t) dt ≤ kwk[a,b],∞ 2 (b − a) " (b − a)(x − a) 3 3 + (b − x)4−(x − a)4 4 ! f 00 (a) + (x − a)4−(b − x)4 4 + (b − a) (b − x)3 3 ! f 00 (b) #
which was given by Sarikaya and Yaldiz in [25].
Corollary 2.13. Under the same assumptions of Theorem 2.9 with w(u)= 1, then we have the inequality n X k=0 (b − x)k+1+ (−1)k(x − a)k+1 (k+ 1)! f (k)(x) − b Z a f(t) dt ≤" (x − a) n+2 (n+ 2)! + (b − x)n+3−(x − a)n+3 (b − a) (n+ 1)! (n + 3) ! f (n+1) (a) + (x − a)n+3−(b − x)n+3 (b − a) (n+ 1)! (n + 3) + (b − x)n+2 (n+ 2)! ! f (n+1) (b) # .
Corollary 2.14. If we take w(u)= 1 and n = 0 in Theorem 2.9, then we have (b − a) f (x) − b Z a f(t) dt ≤ " (x − a) 2 2 + (b − x)3−(x − a)3 3 (b − a) ! f 0 (a) + (x − a)3−(b − x)3 3 (b − a) + (b − x)2 2 ! f 0 (b) # .
Remark 2.15. If we take w(u)= 1 and n = 1 in Theorem 2.9, then we get (b − a) f (x)+ (b − a) a+ b 2 −x ! f0(x) − b Z a f(t) dt ≤" (x − a) 3 6 + (b − x)4−(x − a)4 8 (b − a) ! f 00 (a) + (x − a)4− (b − x)4 8 (b − a) + (b − x)3 6 ! f 00 (b) #
Remark 2.16. If we choose w(u)= 1 and x = a+b2 in Theorem 2.9, then we have the inequality n X k=0 (b − a)k+1h1+ (−1)ki 2k+1(k+ 1)! f (k) a+ b 2 ! − b Z a f(t) dt ≤ (b − a) n+2 2n+1(n+ 2)! f (n+1) (a) + f (n+1) (b) 2 which was derived by Ozdemir and Yildiz in [22].
Remark 2.17. If we take w(u) = 1, x = a+b2 and n = 0 in Theorem 2.9, then the inequality (11) reduce to the inequality (4).
Remark 2.18. If we take w(u)= 1, x = a+b2 and n= 1 in Theorem 2.9, then the inequality (11) becomes the inequality (6).
We prove some inequalities by using convexity of f(n+1) q .
Theorem 2.19. Suppose that all the assumptions of Lemma 2.1 hold. If f
(n+1)
q
is convex on [a, b], q > 1, then, for all x ∈ [a, b] , we have the inequality
n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt (14) ≤ kwk[a,b],∞ (n+ 1)! (b − a) 1 q" (b − x) (n+1)p+1+ (x − a)(n+1)p+1 (n+ 1)p + 1 #1p f(n+1)(a) q + f(n+1)(b) q 2 1 q where 1 p+ 1 q = 1, and kwk∞= sup t∈[a,b] |w(t)|.
Proof. By similar methods in the proof of Theorem 2.9 and from H ¨older’s inequality, we find that n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt ≤ b Z a |Pw(x, t)|pdt 1 p b Z a f (n+1)(t) q dt 1 q . (15)
By simple calculations, we obtain b Z a |Pw(x, t)|pdt 1 p = 1 n! x Z a t Z a (t − u)nw(u)du p dt+ b Z x b Z t (u − t)nw(u)du p dt 1 p (16) ≤ kwk[a,b],∞ (n+ 1)! x Z a (t − a)(n+1)pdt+ b Z x (b − t)(n+1)pdt 1 p = kwk[a,b],∞ (n+ 1)! " (b − x)(n+1)p+1+ (x − a)(n+1)p+1 (n+ 1)p + 1 #1p . Since f (n+1) (t) q
is convex on [a, b] = [a, x] ∪ [x, b], we have f(n+1) b − t b − aa+ t − a b − ab ! q ≤ b − t b − a f (n+1)(a) q + t − a b − a f (n+1)(b) q . (17)
Using the inequality (17), it follows that b Z a f (n+1)(t) q dt 1 q ≤(b − a)1q f(n+1)(a) q + f(n+1)(b) q 2 1 q . (18)
Hence, the proof of theorem is completed.
Corollary 2.20. Under the same assumptions of Theorem 2.19 with n= 0, then the following inequality hols:
f(x) b Z a w(t) dt − b Z a w(t) f (t) dt ≤ kwk[a,b],∞(b − a)1q " (b − x) p+1+ (x − a)p+1 p+ 1 #1p f 0 (a) q + f 0 (b) q 2 1 q
which is ”weighted Ostrowski” inequality provided that f 0 q is convex on [a, b] .
Corollary 2.21. Under the same assumptions of Theorem 2.19 with n= 0 and x = a+b2 , then the following inequality hols: f a+ b 2 !Zb a w(t) dt − b Z a w(t) f (t) dt ≤ kwk[a,b],∞(b − a)2 2 · p+ 11p f 0 (a) q + f 0 (b) q 2 1 q
which is ”weighted mid-point” inequality provided that f
0
is convex on [a, b] .
Remark 2.22. If we choose n= 1 in Theorem 2.19, then we obtain
M1(x) f0(x)+ M0(x) f (x) − b Z a w(t) f (t) dt ≤ kwk[a,b],∞ 2 (b − a) 1 q" (b − x) 2p+1+ (x − a)2p+1 2p+ 1 #1p f 00 (a) q + f 00 (b) q 2 1 q
which was given by Sarikaya and Yaldiz in [25].
Corollary 2.23. Under the same assumptions of Theorem 2.19 with w(u)= 1, then we have the inequality n X k=0 (b − x)k+1+ (−1)k(x − a)k+1 (k+ 1)! f (k)(x) − b Z a f(t) dt ≤ (b − a) 1 q (n+ 1)! " (b − x)(n+1)p+1+ (x − a)(n+1)p+1 (n+ 1)p + 1 #1p f (n+1)(a) q + f (n+1)(b) q 2 1 q .
Remark 2.24. If we take w(u)= 1 and n = 0 in Theorem 2.19, then we get (b − a) f (x) − b Z a f(t) dt ≤ (b − a)1q " (b − x) p+1+ (x − a)p+1 p+ 1 #1p f 0 (a) q + f 0 (b) q 2 1 q .
Remark 2.25. If we take w(u)= 1 and n = 1 in Theorem 2.19, then we have (b − a) f (x)+ (b − a) a+ b 2 −x ! f0(x) − b Z a f(t) dt ≤ (b − a) 1 q 2 " (b − x)2p+1+ (x − a)2p+1 2p+ 1 #1p f 00 (a) q + f 00 (b) q 2 1 q
which was given by Sarikaya and Yaldiz in [25].
Corollary 2.26. If we choose w(u)= 1 and x = a+b2 in Theorem 2.19, then we have the inequality n X k=0 (b − a)k+1h1+ (−1)ki 2k+1(k+ 1)! f (k) a+ b 2 ! − b Z a f(t) dt ≤ (b − a) n+2 2n+1(n+ 1)! (n + 1)p + 11p f (n+1)(a) q + f (n+1)(b) q 2 1 q .
Remark 2.27. If we take w(u)= 1, x = a+b2 and n= 0 in Theorem 2.19, then we have (b − a) f a+ b 2 ! − b Z a f(t) dt ≤ (b − a) 2 2 p+ 11p f 0 (a) q + f 0 (b) q 2 1 q .
Remark 2.28. If we take w(u) = 1, x = a+b2 and n = 1 in Theorem 2.19, then the inequality (14) becomes the inequality (7).
Theorem 2.29. Suppose that all the assumptions of Lemma 2.1 hold. If f
(n+1)
q
is convex on [a, b], q > 1, then for all x ∈ [a, b] , we have the inequality
n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt (19) ≤ kwk[a,b],∞ (b − a)1q (n+ 1)! (n + 1) p + 1 1 p × (x − a)n+1+1p" (b − a) 2− (b − x)2 2 f (n+1) (a) q +(x − a) 2 2 f (n+1) (b) q#1q + (b − x)n+1+1 p" (b − x) 2 2 f (n+1) (a) q +(b − a)2−(x − a)2 2 f (n+1) (b) q#1q where 1p+1q = 1, and kwk∞= sup
t∈[a,b]
|w(t)|.
Proof. Using similar methods in the proof of Theorem 2.19 and from H ¨older’s inequality, we obtain the inequality (19). Hence, the proof is completed.
Remark 2.30. Under the same assumptions of Theorem 2.29 with n= 0, then the following inequality holds: f(x) b Z a w(t) dt − b Z a w(t) f (t) dt ≤ kwk[a,b],∞ (b − a)1q p+ 11p (x − a)1+1p" (b − a) 2− (b − x)2 2 f 0 (a) q +(x − a)2 2 f 0 (b) q# 1 q + (b − x)1+1 p" (b − x) 2 2 f 0 (a) q +(b − a)2−(x − a)2 2 f 0 (b) q# 1 q which is ”weighted Ostrowski” inequality provided that
f
0
q
is convex on [a, b] . This inequality was given by Sarikaya and Erden in [26].
Remark 2.31. Under the same assumptions of Theorem 2.29 with n= 0 and x = a+b2 , then the following inequality hols: f a+ b 2 !Zb a w(t) dt − b Z a w(t) f (t) dt ≤ (b − a) 2 4p+ 11p 3 f 0 (a) q + f 0 (b) q 4 1 q + f 0 (a) q + 3 f 0 (b) q 4 1 q kwk[a,b],∞
which is ”weighted mid-point” inequality provided that f
0
is convex on [a, b] . This inequality was given by Sarikaya and Erden in [26].
Remark 2.32. If we choose n= 1 in Theorem 2.29, then we obtain
M1(x) f0(x)+ M0(x) f (x) − b Z a w(t) f (t) dt ≤ kwk[a,b],∞ 2 (b − a)1q 2p+ 1 1 p (x − a)2+1p" (b − a) 2− (b − x)2 2 f 00 (a) q +(x − a) 2 2 f 00 (b) q# 1 q + (b − x)2+1 p" (b − x) 2 2 f 00 (a) q +(b − a)2−(x − a)2 2 f 00 (b) q# 1 q which was established by Sarikaya and Yaldiz in [25].
Corollary 2.33. Under the same assumptions of Theorem 2.29 with w(u)= 1, then we have the inequality n X k=0 (b − x)k+1+ (−1)k(x − a)k+1 (k+ 1)! f (k)(x) − b Z a f(t) dt ≤ 1 (b − a)1q (n+ 1)! (n + 1) p + 11p × (x − a)n+1+1p" (b − a) 2−(b − x)2 2 f (n+1) (a) q +(x − a)2 2 f (n+1) (b) q#1q + (b − x)n+1+1 p" (b − x) 2 2 f (n+1) (a) q +(b − a)2−(x − a)2 2 f (n+1) (b) q#1q .
Remark 2.34. If we take w(u)= 1 and n = 1 in Theorem 2.29, then we get (b − a) f (x)+ (b − a) a+ b 2 −x ! f0(x) − b Z a f(t) dt ≤ 1 2 (b − a)1q 2p+ 1 1 p (x − a)2+1p" (b − a) 2− (b − x)2 2 f 00 (a) q +(x − a) 2 2 f 00 (b) q# 1 q + (b − x)2+1 p" (b − x) 2 2 f 00 (a) q + (b − a)2−(x − a)2 2 f 00 (b) q# 1 q which was given by Sarikaya and Yaldiz in [25].
Remark 2.35. If we choose w(u)= 1 and x = a+b2 in Theorem 2.29, then we have the inequality n X k=0 (b − a)k+1h1+ (−1)ki 2k+1(k+ 1)! f(k) a+ b 2 ! − b Z a f(t) dt ≤ (b − a) n+2 2n+2(n+ 1)! (n + 1) p + 11p 3 f (n+1) (a) q + f (n+1) (b) q 4 1 q + f (n+1) (a) q + 3 f (n+1) (b) q 4 1 q which was proved by Ozdemir and Yildiz in [22].
Remark 2.36. If we take w(u) = 1, x = a+b2 and n = 0 in Theorem 2.29, then the inequality (19) reduce to the inequality (5).
Remark 2.37. If we take w(u)= 1, x = a+b2 and n= 1 in Theorem 2.29, then we have the inequality f a+ b 2 ! − 1 b − a b Z a f(t) dt ≤ (b − a) 2 162p+ 11p 3 f 00 (a) q + f 00 (b) q 4 1 q + f 00 (a) q + 3 f 00 (b) q 4 1 q which was given by Sarikaya and Yaldiz in [25].
Theorem 2.38. Suppose that all the assumptions of Lemma 2.1 hold. If f(n+1)
q
is convex on [a, b], q ≥ 1, then, for all x ∈ [a, b] , the following inequality holds:
n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt (20) ≤ kwk[a,b],∞ (n+ 1)! (b − a)1q (b − x)n+2+ (x − a)n+2 (n+ 2) !1p × " (b − a)(x − a) n+2 n+ 2 + (b − x)n+3−(x − a)n+3 n+ 3 ! f (n+1) (a) q + (x − a)n+3−(b − x)n+3 n+ 3 + (b − a) (b − x)n+2 n+ 2 ! f (n+1) (b) q# 1 q where 1 p+1q = 1 and kwk∞ = sup t∈[a,b] |w(t)|.
Proof. From (10), using the properties of modulus and from H ¨older’s inequality, we get n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt (21) ≤ b Z a |Pw(x, t)| dt 1 p b Z a |Pw(x, t)| f (n+1)(t) q dt 1 q .
By simple calculations, we obtain
b Z a |Pw(x, t)| dt ≤ kwk[a,b],∞(b − x) n+2+ (x − a)n+2 (n+ 2)! . (22) Because of convexity of f (n+1) q
and bounded of w, appliying similar methods in the proof of Theorem 2.9 and using the inequality (17), we find that
b Z a |Pw(x, t)| f(n+1)(t) q dt (23) ≤ kwk[a,b],∞ (+1)! (b − a) " (b − a)(x − a) n+2 n+ 2 + (b − x)n+3−(x − a)n+3 n+ 3 ! f (n+1) (a) q (x − a)n+3−(b − x)n+3 n+ 3 + (b − a) (b − x)n+2 n+ 2 ! f (n+1) (b) q# .
Substituting the inequalities (22) and (23) in (21), we easily deduce required inequality (20) which completes the proof.
Remark 2.39. In case(p, q) = (∞, 1), if we take limit as p → ∞ in Theorem 2.38, then the inequality (20) becomes
the inequality (11). Thus, we obtain all of the results which are similar to Theorem 2.9.
Remark 2.40. If we take w(u)= 1, x = a+b2 and n= 0 in Theorem 2.38, then we obtain f a+ b 2 ! − 1 b − a b Z a f(t) dt ≤b − a 4 f 0 (a) q + f 0 (b) q 2 1 q
which was established by Pearce and Peˇcari´c in [21].
3. Some applications for the moments
Distribution functions and density functions provide complete descriptions of the distribution of probal-ity for a given random variable. However, they do not allow us to easily make comparisons between two different distributions. The set of moments that uniquely characterizes the distribution under reasonable conditions are useful in making comparisons. Knowing the probability function, we can determine mo-ments if they exist. Applying the mathematical inequalities, some estimations for the momo-ments of random variables were recently studied (see, [3]-[5], [12]-[15], [25]).
Set X to denote a random variable whose probability function is w : [a, b] → R is a integrable and nonnegative function on the interval of real numbers I and let a, b ∈ I, (a < b). Denote by Mr(x) the rth
moment about any arbitrary point x of the random variable X, r ≥ 0, defined as
Mr(x)= b
R
a
(u − x)rw(u) du, r = 0, 1, 2, ...
Now, we reconsider the identity (8) by changing conditions given in Lemma 2.1. Herewith, we deduce an identity involving rthmoment.
Lemma 3.1. Let f : I ⊂ R → R be (n + 1) −times differentiable function on I◦, a, b ∈ I◦ with a < b, f(n+1) is absolutely continuous on [a, b] and let X be random variable whose probability function is w : [a, b] → [0, ∞) is integrable on [a, b]. Then the following equality holds:
n X k=0 Mk(x) k! f (k)(x) − b Z a w(t) f (t) dt= b Z a Pw(x, t) f(n+1)(t) dt
where n ∈ N, Mk(x) is the kth moment and Pw(x, t) is defined as (9).
Similarly, using boundedness of f(n+1), convexity of f (n+1) or convexity of f (n+1) q in addition to condi-tions of Lemma 3.1, we obtain same of the inequalities given in previous section for random variable.
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